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Hardness Amplification Generic Amplification Theorem: If there are problems in class A that are mildly hard for algorithms in Z, then there are problems in A that are very hard for Z. NP, EXP, PSPACE P/poly, BPP, P

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PSPACE versus P/poly, BPP Long line of work: Theorem: If there are problems in PSPACE that are worst case hard for P/poly (BPP), then there are problems that are ½ + hard for P/poly(BPP). Yao, Nisan-Wigderson, Babai-Fortnow-Nisan-Wigderson, Impagliazzo, Impagliazzo-Wigderson1, Impagliazzo- Wigderson2, Sudan-Trevisan-Vadhan, Trevisan-Vadhan, Impagliazzo-Jaiswal-Kabanets, Impagliazzo-Jaiswal-Kabanets- Wigderson.

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NP versus P/poly ODonnell. Theorem: If there are problems in NP that are 1 - hard for P/poly, then there are problems that are ½ + hard. Starts from average-case assumption. Healy-Vadhan-Viola.

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NP versus BPP Trevisan03. Theorem: If there are problems in NP that are 1 - hard for BPP, then there are problems that are ¾ + hard.

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NP versus BPP Trevisan05. Theorem: If there are problems in NP that are 1 - hard for BPP, then there are problems that are ½ + hard. BureshOppenheim-Kabanets-Santhanam: alternate proof via monotone codes. Optimal up to.

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Our results Amplification against P. Theorem 1: If there is a problem in NP that is 1 - hard for P, then there is a problem which is ¾ + hard. Theorem 2: If there is a problem in PSPACE that is1 - hard for P, then there is a problem which is ¾ + hard. Trevisan: 1 - hardness to 7/8 + for PSPACE. Goldreich-Wigderson: Unconditional hardness for EXP against P. = 1/n 100 = 1/(log n) 100

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Beyond Unique Decoding… 100110011100110011 Deterministic local list-decoder: Set L of machines such that: - For any received word - Every nearby codeword is computed by some M 2 L. Is this possible?